Question

Out of 8 numbers of which 4 are odd & 4 are even 2 numbers are selected at random what is the probability that their sum is even?​

Answer 1

here o is odd and e is even

o1,e1,o2,e2,o3,e3,o4,e4

all possible outcomes -

o1 + one of any other , so o1 will give us 7 possible outcomes

similarly all 7 remaining will also give 7 possible outcomes

therefore 8*7 = 56 are total outcomes

here if odd and even comes then the sum will be odd , else the sum will be even

in case of o1 we get 3 favourable outcomes

similarly in all cases we get 3 favourable outcomes where sum is even

so 8*3 is the no. of favourable outcomes= 24

probability = no. of favourable outcomes / no. of total outcomes

=> probability = 24/56 = 3/7

Answer 2

$probability\: of\: pair\: of\: numbers \:with\: sum\: as\: even\: is \frac{3}{7}$

Question :-

Out of 8 numbers of which 4 are odd numbers and 4 are even numbers. If 2 numbers are selected at random what is the probability that their sum is even ?

Given :-

We have 8 numbers out of which 4 is even number and 4 is odd number

To Find :-

probability of the pair of numbers that their sum is even

Formula Applied :-

$probability = \frac{favorable \: outcome}{total \: outcomes}$

Solution :-

Let us assume 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 as 8 numbers out of which 1 , 3 , 5 , 7 as 4 odd numbers and 2 , 4 , 6 , 8 as 4 even numbers.

( NOTE : Its given as 2 numbers are taken as pair so same numbers cant be written as a pair in sample space ).

Sample space = { (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (1,7) , (1,8)

(2,1) , (2,3) , (2,4) , (2,5) , (2,6) , (2,7) , (2,8)

(3,1) , (3,2) , (3,4) , (3,5) , (3,6) , (3,7) , (3,8)

(4,1) , (4,2) , (4,3) , (4,5) , (4,6) , (4,7) , (4,8)

(5,1) , (5,2) , (5,3) , (5,4) , (5,6) , (5,7) , (5,8)

(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,7) , (6,8)

(7,1) , (7,2) , (7,3) , (7,4) , (7,5) , (7,6) , (7.8)

(8,1) , (8,2) , (8,3) , (8,4) , (8,5) , (8,6) , (8,7) }

hence , Total no . of outcomes [ n(s) ] = 56

We know that , Odd + Odd = Even

Even + Even = Even

So the pairs with both odd numbers and with both even numbers should taken as favorable outcome . Let the favorable outcome be as A

Favorable outcomes = { (1,3) , (1,5) , (1,7)

(2,4) , (2,6) , (2,8)

(3,1) , (3,5) , (3,7)

(4,2) , (4,6) , (4,8)

(5,1) , (5,3) , (5,7)

(6,2) , (6,4) , (6,8)

(7,1) , (7,3) , (7,7)

(8,2) , (8,4) , (8,6) }

No . Of Favorable outcomes [ n(A) ] = 24

$probability \: of \: pair \: of \: numbers \: with \: sum \: as \: even = \frac{no.of \: favorable \: outcome \: ({n(A)})}{no.of \: total \: outcomes \: (n(S))}$

$= \frac{24}{56} = \frac{3}{7}$